Question: Let $f(x)=\sqrt{2x+1}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $4\leq x\leq12$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1.5$ (Choice C) C $4$ (Choice D) D $7.5$
Explanation: According to the Mean Value Theorem, there exists a number $c$ in the open interval $4<x<12$ such that $f'(c)$ is equal to the average rate of change of $f$ over the interval: $f'(c)=\dfrac{f(12)-f(4)}{(12)-(4)}$ First, let's find that average rate of change: $\dfrac{f(12)-f(4)}{(12)-(4)}=\dfrac{5-3}{8}={\dfrac{1}{4}}$ Now, let's differentiate $f$ and find the $x$ -value for which $f'(x)={\dfrac{1}{4}}$. $f'(x)=\dfrac{1}{\sqrt{2x+1}}$ The solution of $f'(x)=\dfrac{1}{4}$ is $x=7.5$. $x=7.5$ is indeed within the interval $4<x<12$. In conclusion, $c=7.5$.